D dochelay New member Joined Sep 30, 2010 Messages 1 Sep 30, 2010 #1 Show that if p is prime and 0 < k < p, then (p-k)!(k-1)! = \(\displaystyle (-1)^k\) (mod p) I'm guessing this has something to do with Wilson's Theorem, but I'm having trouble manipulating it.
Show that if p is prime and 0 < k < p, then (p-k)!(k-1)! = \(\displaystyle (-1)^k\) (mod p) I'm guessing this has something to do with Wilson's Theorem, but I'm having trouble manipulating it.
D daon Senior Member Joined Jan 27, 2006 Messages 1,284 Sep 30, 2010 #2 (p-1)! = (p-1)(p-2)...(p-k+1)(p-k)! Taken mod p, the rhs is equal to (0-1)(0-2)...(0-k+1) = (-1)(-2)(-3)...(-k+1)(p-k)! Try to figure it out from here...
(p-1)! = (p-1)(p-2)...(p-k+1)(p-k)! Taken mod p, the rhs is equal to (0-1)(0-2)...(0-k+1) = (-1)(-2)(-3)...(-k+1)(p-k)! Try to figure it out from here...
